Take the expression $(1-(1-\frac{1}{n})^x)$, which is the expected fraction of elements one samples from a set of $n$ elements, sampling with replacement (and uniform probability) $x$ times.

Provided some $n$, how do we analytically solve for $x$ s.t. $(1-(1-\frac{1}{n})^x) \geq P$? It appears that $\frac{x}{n}$ converges to a fixed value as $n \to \inf$. Is this true? If so, how do we find it without relying on numerical methods?

Your expression $(1-(1-\frac{1}{n})^x)$ is always less than $1$, so how can this be the expected number of unique elements?? Also, to even have the slightest nonzero chance of covering a portion $P$ with $x$ samples one needs $\frac xn\geq P$, and this requires $x\to\infty$ as $n\to\infty$. So obviously $x$ cannot.
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Marc van LeeuwenOct 23 '12 at 13:27

@Marc von Leeuwen Sorry, I meant that the expression was the fraction of a set of n elements covered after sampling with replacement x times.
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PlatiOct 23 '12 at 14:17

The inequality can be rewritten as
$$\left(1-\frac{1}{n}\right)^x\le 1-P.$$
We first solve the equation
$$\left(1-\frac{1}{n}\right)^t= 1-P.$$
Take the (natural) logarithm of both sides.
We get
$$t\log\left(1-\frac{1}{n}\right)=\log(1-P).$$
Thus
$$t=\frac{\log(1-P)}{\log\left(1-\frac{1}{n}\right)}.$$
Presumably we want $x$ to be an integer. The smallest integer $x$ for which our inequality holds is $\lceil t\rceil$.

It is clear that $x\to \infty$ as $n\to\infty$. However, something interesting happens if we look at $\frac{x}{n}$. We look instead at the closely related $\frac{t}{n}$, which has the same limit. We have
$$\frac{t}{n}=\frac{\log(1-P)}{n\log\left(1-\frac{1}{n}\right)}.$$
Note that the denominator $n\log\left(1-\frac{1}{n}\right)$ is equal to
$\log\left(\left(1-\frac{1}{n}\right)^n\right)$. But $(1-1/n)^n\to e^{-1}$ as $n\to\infty$, so the denominator approaches $-1$. It follows that as $n\to\infty$, $\dfrac{t}{n}$ (and therefore $\dfrac{x}{n}$) approaches $-\log(1-P)$.